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Proof step 30 from proof key-set-write-nodups (previous step)

kar[m] = key-index(key, m2), kar[m1] = key-index(key0, m3), keyset(kar[n, ki], m0) = (keyset(kar, m0) -- kar[n]) ++ ki, n < m0, m0 < # kar, n < # kar, n + 1 < # kar, ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m].idx ≠ kar[m1].idx ⊦ m0 = n, ∀ ki0. ki0 ∈ (keyset(kar, m0) -- kar[n]) ++ kar[m0] ++ ki ↔ ki0 = ki ∨ ki0 ∈ (keyset(kar, m0) ++ kar[m0]) -- kar[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
kar[m1] = key-index(key0, m3), kar[m] = key-index(key, m2), keyset(kar[n, ki], m0) = (keyset(kar, m0) -- kar[n]) ++ ki, n + 1 < # kar, n < # kar, m0 < # kar, n < m0, ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m].idx ≠ kar[m1].idx, ki0 = kar[m0] ∨ ki0 ≠ kar[n] ∧ ki0 ∈ keyset(kar, m0) ⊦ ki0 = ki, m0 = n, ki0 ≠ kar[n] ∧ (ki0 = kar[m0] ∨ ki0 ∈ keyset(kar, m0))

Rule argument

Additional information

Used simplifier rules: ⊦ ki ∈ s -- ki0 ↔ ki ≠ ki0 ∧ ki ∈ s; (s)% ⊦ ki ∈ s ++ ki0 ↔ ki = ki0 ∨ ki ∈ s; (s)